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Solve the absolute value equation: | 1 4 x | + 8 = 13.

x = −1 , x = 3 2

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Solving other types of equations

There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form, and rational equations that result in a quadratic.

Solving equations in quadratic form

Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include x 4 5 x 2 + 4 = 0 , x 6 + 7 x 3 8 = 0 , and x 2 3 + 4 x 1 3 + 2 = 0. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.

Quadratic form

If the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form , which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.

Given an equation quadratic in form, solve it.

  1. Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.
  2. If it is, substitute a variable, such as u , for the variable portion of the middle term.
  3. Rewrite the equation so that it takes on the standard form of a quadratic.
  4. Solve using one of the usual methods for solving a quadratic.
  5. Replace the substitution variable with the original term.
  6. Solve the remaining equation.

Solving a fourth-degree equation in quadratic form

Solve this fourth-degree equation: 3 x 4 2 x 2 1 = 0.

This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let u = x 2 . Rewrite the equation in u .

3 u 2 2 u 1 = 0

Now solve the quadratic.

3 u 2 2 u 1 = 0 ( 3 u + 1 ) ( u 1 ) = 0

Solve each factor and replace the original term for u.

3 u + 1 = 0 3 u = −1 u = 1 3 x 2 = 1 3 x = ± i 1 3
u 1 = 0 u = 1 x 2 = 1 x = ±1

The solutions are ± i 1 3 and ± 1.

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Solve using substitution: x 4 8 x 2 9 = 0.

x = −3 , 3 , i , i

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Solving an equation in quadratic form containing a binomial

Solve the equation in quadratic form: ( x + 2 ) 2 + 11 ( x + 2 ) 12 = 0.

This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting u = x + 2. Then rewrite the equation in u.

u 2 + 11 u 12 = 0 ( u + 12 ) ( u 1 ) = 0

Solve using the zero-factor property and then replace u with the original expression.

u + 12 = 0 u = −12 x + 2 = −12 x = −14

The second factor results in

u 1 = 0 u = 1 x + 2 = 1 x = −1

We have two solutions: −14 , and −1.

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Solve: ( x 5 ) 2 4 ( x 5 ) 21 = 0.

x = 2 , x = 12

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Solving rational equations resulting in a quadratic

Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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